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$\text{Let }\dfrac{x}{y}=z.\\ 2 < \dfrac{x - y}{x + y} < 5\\ 2 < \dfrac{z - 1}{z + 1} < 5\\ 2z + 2 \stackrel{\boxed{1}}{<} z - 1 \stackrel{\boxed{2}}{<} 5z + 5\\ \boxed{1}: 2z + 2 < z - 1\\ z < -3\\ \boxed{2}: z - 1 < 5z + 5\\ -6 < 4z\\ z > -\dfrac{3}{2}\\ \text{Combine }\boxed{1} \text{ and }\boxed{2}: -\dfrac{3}{2} < z < -3\\ \text{The only integer solution is } z = -2\\ \therefore \dfrac{x}{y} = -2$

Note that x-y must be larger than x+y if the ratio is to be between 2 and 5. This will be the case if y is negative.

So we have (x-y)/(x+y) > 2,  or. x-y > 2x+2y or x < -3y or x/y > -3 where the < becomes > because we are dividing by a negative number.

We also have (x-y)/(x+y)<5,  or x-y < 5x+5y or 4x > -6y or x > -(3/2)y or x/y < -3/2 where > becomes < because we are dividing by a negative number.

Hence -3 < x/y < -3/2, giving x/y = -2

Max got the right answer with incorrect reasoning! It happens!

Thank you to the numerous amount of people who answered this question